Problem: Igor the analyst has adopted n little bunnies. As we all know, bunnies love carrots. Thus, Igor has bought a carrot to be shared between his bunnies. Igor wants to treat all the bunnies equally, and thus he wants to cut the carrot into n pieces of equal area. 

Formally, the carrot can be viewed as an isosceles triangle with base length equal to 1 and height equal to h. Igor wants to make n - 1 cuts parallel to the base to cut the carrot into n pieces. He wants to make sure that all n pieces have the same area. Can you help Igor determine where to cut the carrot so that each piece have equal area?

<image> Illustration to the first example.

Input

The first and only line of input contains two space-separated integers, n and h (2 ≤ n ≤ 1000, 1 ≤ h ≤ 105).

Output

The output should contain n - 1 real numbers x1, x2, ..., xn - 1. The number xi denotes that the i-th cut must be made xi units away from the apex of the carrot. In addition, 0 < x1 < x2 < ... < xn - 1 < h must hold. 

Your output will be considered correct if absolute or relative error of every number in your output doesn't exceed 10 - 6.

Formally, let your answer be a, and the jury's answer be b. Your answer is considered correct if <image>.

Examples

Input

3 2


Output

1.154700538379 1.632993161855


Input

2 100000


Output

70710.678118654752

Note

Definition of isosceles triangle: <https://en.wikipedia.org/wiki/Isosceles_triangle>.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int MX = 1e5 + 69;
int n;
long double h, oa;
bool check(long double nx, long double lx) {
  long double lv = (h - lx) / h;
  long double nv = (h - nx) / h;
  long double ar = (nx - lx) * (lv + nv) / 2.0;
  if (ar >= oa) return 1;
  return 0;
}
int main() {
  cin >> n >> h;
  long double lv = 1, lx = 0;
  oa = h / (2.0 * n);
  vector<long double> v;
  for (int i = 1; i < n; i++) {
    long double st = lx, en = h;
    for (int j = 0; j < 123; j++) {
      long double mid = (st + en) / 2.0;
      if (check(mid, lx))
        en = mid;
      else
        st = mid;
    }
    v.push_back(h - st);
    lx = st;
  }
  cout << fixed;
  cout.precision(10);
  for (int i = n - 2; i >= 0; i--) cout << v[i] << ' ';
  return 0;
}